What is the equation of the line that is perpendicular to the given line and passes through the point [tex]\((3, 0)\)[/tex]?

A. [tex]\(3x + 5y = -9\)[/tex]
B. [tex]\(3x + 5y = 9\)[/tex]
C. [tex]\(5x - 3y = -15\)[/tex]
D. [tex]\(5x - 3y = 15\)[/tex]



Answer :

To determine the equation of the line perpendicular to the given line [tex]\( 3x + 5y = -9 \)[/tex] that passes through the point [tex]\((3, 0)\)[/tex], follow these steps:

1. Find the slope of the given line [tex]\(3x + 5y = -9\)[/tex]:
- Start with the standard form [tex]\(Ax + By = C\)[/tex]. The slope of a line [tex]\(Ax + By = C\)[/tex] is given by [tex]\(-A/B\)[/tex].
- For the given line [tex]\(3x + 5y = -9\)[/tex], [tex]\(A = 3\)[/tex] and [tex]\(B = 5\)[/tex].
- Thus, the slope [tex]\(m\)[/tex] of the line is [tex]\(-\frac{3}{5}\)[/tex].

2. Determine the slope of the line perpendicular to the given line:
- The slope of a line perpendicular to another is the negative reciprocal of the slope of the original line.
- The negative reciprocal of [tex]\(-\frac{3}{5}\)[/tex] is [tex]\(\frac{5}{3}\)[/tex].

3. Use point-slope form to find the equation of the perpendicular line:
- Point-slope form is given by [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the given point.
- Substituting [tex]\(m = \frac{5}{3}\)[/tex] and [tex]\((x_1, y_1) = (3, 0)\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \][/tex]
[tex]\[ y = \frac{5}{3}x - \frac{5}{3} \cdot 3 \][/tex]
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]

4. Convert the equation to standard form:
- Multiply both sides of the equation by 3 to eliminate the fraction:
[tex]\[ 3y = 5x - 15 \][/tex]
- Rearrange to form [tex]\(Ax + By = C\)[/tex]:
[tex]\[ 5x - 3y = 15 \][/tex]

5. Check which provided line matches this equation:
- The given options were:
[tex]\[ \begin{aligned} &1. \ 3x + 5y = -9 \\ &2. \ 3x + 5y = 9 \\ &3. \ 5x - 3y = -15 \\ &4. \ 5x - 3y = 15 \\ \end{aligned} \][/tex]
- The equation [tex]\(5x - 3y = 15\)[/tex] matches option 4.

Therefore, the equation of the line that is perpendicular to the given line [tex]\(3x + 5y = -9\)[/tex] and passes through the point [tex]\((3,0)\)[/tex] is given by:

[tex]\[ \boxed{4} \][/tex]

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